Question

1.How do you solve `(1+i)^20-1=80`?

Answer 1

In case `i` is imaginary number, `(1+i)^20-1=-1025`
and not `80`. But if `i` is just a variable `i=0.24573`

Explanation:

`Solution 1` - This assumes `1+i` is a complex number, where `i^2=-1`. In that case `(1+i)^20-1=-1025` and not `80`.
We can write `1+i` in trigonometric form as

`sqrt2(1/sqrt2+i*1/sqrt2)` or `sqrt2(cos45^@+isin45^@)`

and using DeMoivre's theorem, we can write `(1+i)^20` as

`(sqrt2)^20(cos(20xx45^@)+isin(20xx45^@))`

= `2^10(cos900^@+isin900^@)`

= `1024(cos(2xx360^@+180^@)+isin(2xx360^@+180^@))`

= `1024(cos(180^@)+isin(180^@))`

= `1024(-1+i*0)`

= `-1024`

and hence `(1+i)^20-1=-1025`

`Solution 2` - There is another possibility that `i` is a variable and not `i^2=-1`, in that case we can write `(1+i)^20-1=80` as

`(1+i)^20=81`

and `1+i=81^(1/20)=1.24573`

and `i=1.24573-1=0.24573`